The area bounded by `y=x^(2),y=[x+1], 0 le x le 2` and the y-axis is where `[.]` is greatest integer function.

Options

A. `(1)/(3)`
B. `(sqrt(2))/(3)`
C. 1
D. `(7)/(3)`

Explanation

To find the area bounded by the curves \( y = x^2 \), \( y = [x + 1] \) (where \([.]\) denotes the greatest integer function), and the y-axis for \( 0 \leq x \leq 2 \), we will follow these steps:

### Step 1: Understand the Functions
1. **Identify the functions**:
- The function \( y = x^2 \) is a parabola opening upwards.
- The function \( y = [x + 1] \) gives the greatest integer less than or equal to \( x + 1 \). For \( 0 \leq x < 1 \), \( [x + 1] = 1 \). For \( 1 \leq x < 2 \), \( [x + 1] = 2 \).

### Step 2: Determine Intersection Points
2. **Find the intersection points**:
- For \( 0 \leq x < 1 \): \( y = 1 \) intersects \( y = x^2 \).
- Set \( x^2 = 1 \) → \( x = 1 \) (only valid at the boundary).
- For \( 1 \leq x < 2 \): \( y = 2 \) intersects \( y = x^2 \).
- Set \( x^2 = 2 \) → \( x = \sqrt{2} \).

### Step 3: Set Up the Area Calculation
3. **Set up the area calculation**:
- The area can be split into two parts:
- From \( x = 0 \) to \( x = 1 \): Area = \( \int_0^1 (1 - x^2) \, dx \)
- From \( x = 1 \) to \( x = \sqrt{2} \): Area = \( \int_1^{\sqrt{2}} (2 - x^2) \, dx \)

### Step 4: Calculate the First Integral
4. **Calculate the first integral**:
\[ \int_0^1 (1 - x^2) \, dx = \left[ x - \frac{x^3}{3} \right]_0^1 = \left( 1 - \frac{1}{3} \right) - (0) = \frac{2}{3} \]

### Step 5: Calculate the Second Integral
5. **Calculate the second integral**:
\[ \int_1^{\sqrt{2}} (2 - x^2) \, dx = \left[ 2x - \frac{x^3}{3} \right]_1^{\sqrt{2}} \]
- Calculate at \( x = \sqrt{2} \):
\[ 2\sqrt{2} - \frac{(\sqrt{2})^3}{3} = 2\sqrt{2} - \frac{2\sqrt{2}}{3} = \frac{6\sqrt{2}}{3} - \frac{2\sqrt{2}}{3} = \frac{4\sqrt{2}}{3} \]
- Calculate at \( x = 1 \):
\[ 2(1) - \frac{1^3}{3} = 2 - \frac{1}{3} = \frac{6}{3} - \frac{1}{3} = \frac{5}{3} \]
- Therefore, the area from \( 1 \) to \( \sqrt{2} \):
\[ \frac{4\sqrt{2}}{3} - \frac{5}{3} = \frac{4\sqrt{2} - 5}{3} \]

### Step 6: Combine the Areas
6. **Combine the areas**:
\[ \text{Total Area} = \frac{2}{3} + \frac{4\sqrt{2} - 5}{3} = \frac{2 + 4\sqrt{2} - 5}{3} = \frac{4\sqrt{2} - 3}{3} \]

### Final Answer
Thus, the area bounded by the curves is:
\[ \text{Area} = \frac{4\sqrt{2} - 3}{3} \]